Browse Questions

# Verify mean value theorem for each of the function $f(x)=\sqrt {25-x^2}\;in\;[1,5].$

Toolbox:
• Let $f(x)$ be a function defined on $[a,b]$ such that
• (i) it is continuous on $[a,b]$
• (ii) it is differentiable on $(a,b)$
• Then,there exists $c\in (a,b)$ such that $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
$f(x)=\sqrt{25-x^2}$ in $[1,5]$
Since for each $x\in [1,5]$,the function attains a unique definite value.So $f(x)$ is continuous in $[1,5]$
Also differentiable in $(1,5)$
Hence the two conditions of Lagrange's mean value theorem are satisfied.
Hence there must exist some $c\in (1,5)$ such that,
$f'(c)=\large\frac{f(5)-f(1)}{5-1}$
$f'(x)=\large\frac{1}{2\sqrt{25-x^2}}$$(-2x) \qquad=\large\frac{-x}{\sqrt{25-x^2}} Therefore f'(c)=\large\frac{-c}{\sqrt{25-c^2}} f(1)=\sqrt{25-1} \qquad=\sqrt{24} f(5)=\sqrt{25-5} \qquad=0 Step 2: Therefore \large\frac{-c}{\sqrt{25-c^2}}=$$\sqrt{24}$
Squaring on both sides we get,
$\large\frac{c^2}{25-c^2}$$=24$
$c^2=24(25-c^2)$
$c^2=24\times 25-24c^2$
$\Rightarrow 25c^2=24\times 25$
$c^2=24$
$c=2\sqrt 6$
But $\sqrt 6=2.23$
Therefore $c=2\times 2.23$
$\qquad\qquad\;\;=4.46$
This lies in $[1,5]$
Hence Lagrange's mean value theorem is verified.